problem 1 - nd.edupowers/ame.60611/hwex.13/hw4.sol.pdf · examining this problem, we see that as...

Mathematical Methods I - Fall 2013 Homework 4 Solutions

Problem 1Kaplan, p. 436: 2c,i

Find the first three nonzero terms of the following Taylor series:

c) ln (1 + x)2

about x = 0

i) arctanhx about x = 0

The definition of a Taylor Series is:

f(x) =

∞∑n=0

1

n!

∂nf

∂xn

∣∣∣∣x=x0

(x− x0)n (1)

Part c

Using the definition from Equation 1 for (2c), the summary in Table 1 below is produced.

n ∂nf∂xn

∂nf∂xn

∣∣∣x=x0

1n!

∂nf∂xn

∣∣∣x=x0

1n!

∂nf∂xn

∣∣∣x=x0

(x− xo)n

0 log(1 + x)2 0 0 0

1 2 log(1+x)1+x 0 0 0

2 2(1+x)2 −

2 log(1+x)(1+x)2 2 1 x2

3 − 6(1+x)3 + 4 log(1+x)

(1+x)3 −6 −1 −x3

4 22(1+x)4 −

12 log(1+x)(1+x)4 22 11

121112x

4

Table 1: Development of Taylor Series for ln (x+ 1)2

Summing the elements of the final column, the three term approximation is:

f ≈ x2 − x3 +11

12x4

Part i

Using the definition from Equation 1 for (2c), the summary in Table 2 below is produced.

n ∂nf∂xn

∂nf∂xn

∣∣∣x=x0

1n!

∂nf∂xn

∣∣∣x=x0

1n!

∂nf∂xn

∣∣∣x=x0

(x− xo)n

0 arctanhx 0 0 0

1 11−x2 1 1 x

2 2x(1−x2)2 0 0 0

3 8x2

(1−x2)3 + 2(1−x2)2 2 1

3x3

3

4 48x3

(1−x2)4 + 24x(1−x2)3 0 0 0

5 384x4

(1−x2)5 + 288x2

(1−x2)4 + 24(1−x2)3 24 1

5x5

5

Table 2: Development of Taylor Series for arctanhx

Summing the elements of the final column, the three term approximation is:

f ≈ x+x3

3+x5

5


Problem 2Kaplan, p. 649: 5

Show that

J0(x) =

∞∑n=0

(−1

4

)nx2n

(n!)2(2)

Satisfies Bessel’s equation of order 0 in the form

xy′′ + y′ + xy = 0 (3)

Computing the fist and second derivatives of Equation 2 as

J ′0(x) =

∞∑n=1

2n

(−1

4

)nx2n−1/(n!)2

J ′′0 (x) =

∞∑n=1

2n(2n− 1)

(−1

4

)nx2n−2/(n!)2

Remapping these indices of the summation to n = 0

J ′0(x) =

∞∑n=0

2(n+ 1)

(−1

4

)(−1

4

)nx2n+1/((n+ 1)!)2

J ′′0 (x) =

∞∑n=0

2(n+ 2)(2n+ 1)

(−1

4

)(−1

4

)nx2n/((n+ 1)!)2 (4)

Placing the definitions from Equations 2 and 4, into Equation 3 and combining like terms,

∞∑n=0

[{(2n+ 2)(2n+ 1)

(−1

4

)+ 2(n+ 1)

(−1

4

)+ (n+ 1)2

}(−1

4

)nx2n+1/(n+ 1)!

]= 0 (5)

For this to be true for the summation it must hold for all values of x and n, therefore

(2n+ 2)(2n+ 1)

(−1

4

)+ 2(n+ 1)

(−1

4

)+ (n+ 1)2 = 0 (6)

And upon expanding Equation 6, we see that this is true and therefore the series from Equation 2 satisfies Equation3.


Problem 3Course notes, 4.1a

Solve as a series in x for x > 0 about the point x = 0:

x2y′′ − 2xy′ + (x+ 1)y = 0

y(1) = 1 (7)

y(4) = 0

Examining Equation 7, and recognizing that is of the form

P (x)y′′ +Q(x)y′ +R(x)y = 0

Since P (0) = 0, is not an ordinary point. Examining xQ(x)/P (x) = −2 and x2Q(x)/P (x) = x+ 1 are both analyticat x = 0 therefore, this is a regular singular point problem. Therefore there exist a solution for y is of the form:

y =

∞∑n=0

anxn+r (8)

Plugging this assumed form into the differential equation and simplifying yields,[ ∞∑n=0

an(n+ r)(n+ r − 1)xn+r

]+

[ ∞∑n=0

−2an(n+ r)xn+r

]+

[ ∞∑n=0

anxn+r+1

]+

[ ∞∑n=0

anxn+r

]= 0

Combining like powers of x into a single summation and simplifying the result[ ∞∑n=0

an[(n+ r)2 − 3(n+ r) + 1]xn+r

]+

[ ∞∑n=0

anxn+r+1

]= 0

Removing the n = 0 term from the first summation and remapping the index back to zero,

a0(r2 − 3r + 1)xr +

[ ∞∑n=0

an+1[(n+ r + 1)2 − 3(n+ r + 1) + 1]xn+r+1

]+

[ ∞∑n=0

anxn+r+1

]= 0

Combining the two summations

a0(r2 − 3r + 1)xr +

∞∑n=0

[an + an+1{(n+ r + 1)2 − 3(n+ r + 1) + 1}]xn+r+1 = 0

The term outside the summation gives us the indicial equation

r2 − 3r + 1 = 0

Therefore r = 3±√5

2 . Using one of these values, the term outside of the summation is zero and therefore the coefficientof xn+r+1 inside the summation must be zero for all values of x and n. Therefore,

an + an+1((n+ r + 1)2 − 3(n+ r + 1) + 1) = 0

an+1 = − an(n+ r + 1)2 − 3(n+ r + 1) + 1

For r = 3+√5

2

an+1 = − an

(n+ 1 +√

5)(n+ 1)= − a0

(n+ 1)!n+1∏i=1

(i+√

5)

Therefore, the first solution to Equation 7 is,

y1 = a0x12 (3+

√5)(

1− 1

1 +√

5x+

1

14 + 6√

5x2 +

1

24

(4√

5− 9)x3 + . . .

)


For r = 3−√5

2

bn+1 = − bn

(n+ 1−√

5)(n+ 1)= − b0

(n+ 1)!n+1∏i=1

(i−√

5)

Therefore, the second solution to Equation 7 is,

y2 = b0x12 (3−

√5)

(1 +

1√5− 1

x+1

8

(7 + 3

√5)x2 +

(7 + 3

√5)

24(√

5− 3)x3 + . . .

)Combining the two solutions, y = y1+y2, then considering the boundary conditions a0 = 0.675038 and b0 = 0.183975.Therefore, we can plot the solution below.

1 2 3 4

0.5

1.0

1.5

2.0

2.5

3.0

3.5

Exact

4 Term Approximation

Figure 1: Exact and approximate solution to x2y′′ − 2xy′ + (x+ 1)y = 0


Problem 4Course notes, 4.2

Find two-term expansions for each of the roots of:

(x− 1)(x+ 3)(x− 3λ) + 1 = 0

where λ is large.Multiplying out the terms:

x3 + (2− 3λ)x2 − (6λ+ 3)x+ (1 + 9λ) = 0

Dividing through by λ, and substituting in a small number ε for 1λ ,

εx3 + (2ε− 3)x2 − (6 + 3ε)x+ (ε+ 9) = 0 (9)

Examining this problem, we see that as ε → 0, a solution is lost. Therefore, we must change variables, we canperform the transformation,

X =x

εα(10)

Using the transformation from Equation 10 on Equation 9,

X3ε3α+1 − 3X2ε2α + 2X2ε2α+1 − 3X(ε+ 2)εα + ε+ 9 = 0 (11)

The two highest order terms of X are in the same order of ε if 1 + 3α = 2α, Therefore, we demand that α = −1.With this, Equation 11 becomes,

X3 +X2(2ε− 3)− 3Xε(ε+ 2) + ε2(ε+ 9) = 0 (12)

Let x be expressed as a sum in ε,

X =

∞∑n=0

Xnεn ≈ X0 +X1ε+X2ε

2 +X3ε3 (13)

Therefore, separating on orders of ε and recognizing that since the expansion is 0 each term must also equal 0, theequations become

X30 − 3X2

0 = 0

3X1X20 + 2X2

0 − 6X1X0 − 6X0 = 0

3X2X20 + 3X2

1X0 + 4X1X0 − 6X2X0 − 3X0 − 3X21 − 6X1 + 9 = 0

X31 + 2X2

1 + 6X0X2X1 − 6X2X1 − 3X1 + 4X0X2 − 6X2 + 3X20X3 − 6X0X3 + 1 = 0

Sequentially solving these equations until we have two terms for each solution, we find

(X0, X1, X2) =

(0,−3,− 1

12

)(14)

(X0, X1, X2) =

(0, 1,

1

12

)(X0, X1, X2, X3) =

(3, 0, 0,−1

9

)Then combining the values from Equation 14 and the initial approximation from Equation 13, we find,

X ≈ − ε2

12− 3ε,

ε2

12+ ε or 3− ε3

9

Therefore

x ≈ − ε

12− 3,

ε

12+ 1, or

3

ε− ε2

9

x ≈ − 1

12λ− 3,

1

12λ+ 1, or 3λ− 1

9λ2


Problem 5Course notes, 4.11b

Find all solutions through O(ε2), where ε is a small parameter, and compare with the exact result for ε = 0.01.

2εx4 + 2(2ε+ 1)x3 + (7− 2ε)x2 − 5x− 4 = 0 (15)

Starting this problem, we see that as ε→ 0, a solution is lost. Therefore, we must change variables, we can performthe transformation,

X =x

εα(16)


2ε4α+1X4 + 2(2ε3α+1 + ε3α)X3 + (7ε2α − 2ε2α+1)X2 − 5εαX − 4 = 0 (17)

The two highest order terms of X are in the same order of ε if 4α + 1 = 3α, Therefore, we demand that α = −1.With this, Equation 17 becomes,

2ε−3X4 + 2(2ε−2 + ε−3)X3 + (7ε−2 − 2ε−1)X2 − 5ε−1X − 4 = 0 (18)

In order to solve this problem, we start by assuming that X can be written in the form,

X =∞∑n=0

Xnεn = X0 +X1ε+X2ε

2 +X3ε3 + . . . (19)

Then placing the expansion from Equation 19 into Equation 18, we find,

2ε−3(X0 +X1ε+X2ε2 +X3ε

3 + . . . )4 + 2(2ε−2 + ε−3)(X0 +X1ε+X2ε2 +X3ε

3 + . . . )3+ (20)

(7ε−2 − 2ε−1)(X0 +X1ε+X2ε2 +X3ε

3 + . . . )2 − 5ε−1(X0 +X1ε+X2ε2 +X3ε

3 + . . . )− 4 = 0

Expanding this and separating by powers of ε and recognizing that if the sum of all terms is 0 then each order of εmust be 0 as well. Examining these equations,

ε0 −2X40 − 2X3

0 = 0 (21)

ε1 −8X1X30 − 4X3

0 − 6X1X20 − 7X2

0 = 0

ε2 −8X2X30 − 12X2

1X20 − 12X1X

20 − 6X2X

20 + 2X2

0 − 6X21X0 − 14X1X0 + 5X0 = 0

ε3 −8X3X30 − 24X1X2X

20 − 12X2X

20 − 6X3X

20 − 8X3

1X0 − 12X21X0 + 4X1X0

−12X1X2X0 − 14X2X0 − 2X31 − 7X2

1 + 5X1 + 4 = 0

......

Solving Equations 21 we see,

(X0, X1, X2, X3) =

(0,−4,−32

5,−31232

875

),

(0,−1

2,− 1

12,− 11

378

),

(0, 1,− 4

15,

752

3375

),

(−1,

3

2,

27

4,

71

2

)(22)

Therefore using the definition from Equation 19, the data from Equation 22 and the transformation defined inEquation 16, the O(ε2) solutions are,

X = −4ε− 32

5ε2 − 31232

875ε3 → x = −31232

875ε2 − 32

5ε− 4

X = −1

2ε− 1

12ε2 − 11

378ε3 → x = − 11

378ε2 − 1

12ε− 1

2

X = ε− 4

15ε2 +

752

3375ε3 → x =

752

3375ε2 − 4

15ε+ 1

X = −1 +3

2ε+

27

4ε2 +

71

2ε3 → x =

71

2ε2 +

27

4ε+

3

2− 1

ε

Now, in the case of ε = 0.01, our estimate of the roots of Equation 15 would be

x = −4.06757,−0.500836, 0.997356,−98.429

While the exact solution is,x = −4.06783,−0.500836, 0.997355,−98.4287

The O(ε2) approximation provides an excellent approximation of the roots, the maximum relative error is 6.5×10−5.



Find three terms of a solution ofx+ ε cos (x+ 2ε) =

π

2(23)

where ε is a small parameter. For ε = 0.2, compare the best asymptotic solution with the exact solution.In order to solve this problem, we start by assuming that x can be written in the form,

x =

∞∑n=0

xnεn = x0 + x1ε+ x2ε

2 + x3ε3 + . . . (24)

Then placing the approximation from 24 into Equation 23, we find,

(x0 + x1ε+ x2ε2 + x3ε

3 + . . . ) + ε cos ((x0 + x1ε+ x2ε2 + x3ε

3 + . . . ) + 2ε) =π

2(25)

Performing a Taylor series expansion in ε about ε = 0, splitting this up by powers of ε,

ε0 : x0 −π

2= 0 (26)

ε1 : x1 + cos(x0) = 0

ε2 : x2 − (x1 + 2) sin(x0) = 0

ε3 : x3 − x2 sin(x0)− 1

2(x1 + 2)2 cos(x0) = 0

......

Solving for the terms in Equation 26, we find x0 = π2 , x1 = 0, x2 = 2, x3 = 2. Therefore the three term asymptotic

solution of Equation 23 is,

x =π

2+ 2ε2 + 2ε3

In the case of ε = 0.2, this yields xAppx = 1.6668, while the exact solution is xExact = 1.6658.



The solution of the matrix equation A · x = y can be written as x = A−1 · y. Find the perturbation solution of

(A + εB) · x = y, (27)

where ε is a small parameter.Assuming x is of the form

x =

∞∑n=0

εnxn = x0 + εx1 + ε2x2 + . . . (28)

This means that Equation 27 can be written as,

(A + εB) · (x0 + εx1 + ε2x2 + . . . ) = y (29)

Distributing through the dot product and grouping by powers of ε, Equation 29 becomes,

(A · x0 − y) + ε(B · x0 + A · x1) + ε2(B · x1 + A · x2) + · · · = 0 (30)

Since the summation of the terms is zero, each power of ε must be as well therefore,

A · x0 = y

B · x0 + A · x1 = 0

B · x1 + A · x2 = 0

...

Solving each line for the unknown x,

x0 = A−1 · yx1 = −A−1 ·B · x0

x2 = −A−1 ·B · x1

...

Back substituting, we find that in general,

xn = A−1 ·(−B ·A−1

)n · y (31)

Therefore using the definition of x from Equation 28 and the values of xn from Equation 31, the perturbation solutionof Equation 27 is,

x = A−1 ·

[ ∞∑n=0

(−εB ·A−1

)n] · y



Find all solutions ofεx4 + x− 2 = 0 (32)

approximately, if ε is small and positive. If ε = 0.001, compare the exact solution obtained numerically with theasymptotic solution.Note as ε→ 0, the equation becomes singular. Let

X =x

εα(33)


ε4α+1X4 + εαX − 2 = 0 (34)

The two highest order terms of X are in the same order of ε if 4α + 1 = α, Therefore, we demand that α = − 13 .

With this, Equation 34 becomes,ε−

13X4 + ε−

13X − 2 = 0 (35)

In order to solve this problem, we start by assuming that X can be written in the form,

X =

∞∑n=0

Xnεn3 = X0 +X1ε

13 +X2ε

23 +X3ε+ . . . (36)

Then placing the expansion from Equation 36 into Equation 35, we find,

(X0 +X1ε13 +X2ε

23 +X3ε+ . . . )4 + (X0 +X1ε

13 +X2ε

23 +X3ε+ . . . )− 2ε

13 = 0 (37)

Expanding this and separating by powers of ε and recognizing that if the sum of all terms is 0 then each order of εmust be 0 as well. Examining these equations,

ε0 X40 +X0 = 0 (38)

ε13 4X1X

30 +X1 − 2 = 0

ε23 4X2X

30 + 6X2

1X20 +X2 = 0

ε1 4X3X30 + 12X1X2X

20 + 4X3

1X0 +X3 = 0

......

Solving Equations 38 we see,

(X0, X1, X2, X3) =

(−1,−2

3,

8

9,−160

81

), (39)

(0, 2, 0, 0) ,(3√−1,−2

3,

8

9

(3√−1− 1

),

160 3√−1

81

),(

− (−1)2/3

,−2

3,

8

9

(−1− (−1)

2/3),

1

81(−160) (−1)

2/3

)Therefore using the definition from Equation 36, the data from Equation 39 and the transformation defined inEquation 33, the O(ε) solutions are,

X =8ε2/3

9− 1

32 3√ε− 160ε

81− 1

X = 2 3√ε

X =8

9

(3√−1− 1

)ε2/3 − 1

32 3√ε+

1

813√−1160ε+ 3

√−1

X =8

9

(−1− (−1)2/3

)ε2/3 − 1

32 3√ε− 160

81(−1)2/3ε− (−1)2/3


Performing the inverse transformation to convert X to x,

x = −160ε2/3

81+

8 3√ε

9− 1

3√ε− 2

3

x = 2

x =89

(3√−1− 1

)ε2/3 − 1

32 3√ε+ 1

813√−1160ε+ 3

√−1

3√ε

x =−72

(1 + (−1)2/3

)ε2/3 − 54 3

√ε− 160(−1)2/3ε− 81(−1)2/3

81 3√ε

Using this approximation when ε = 0.001, we find the approximate roots are

x1 = −10.5975

x2 = 2

x3 = 4.29877 + 8.75434i

x4 = 4.29877− 8.75434i

and the exact roots are

x1 = −10.5934

x2 = 1.98449

x3 = 4.30446− 8.75258i

x4 = 4.30446 + 8.75258i



Obtain the first two terms of an approximate solution to

x+ 3(1 + ε)x+ 2x = 0, with (40)

x(0) = 2(1 + ε),

x(0) = −3(1 + 2ε),

for small ε. Compare with the exact solution graphically in the range 0 ≤ t ≤ 1 for (a) ε = 0.1, (b) ε = 0.25 and (c)ε = 0.5.Letting x be of the form x = x0 + εx1 + . . . . Therefore the second derivative takes the form, x = x0 + εx1 + . . . andthe second derivative takes the form, x = x0 + εx1 + . . . . Equation 40 then becomes,

(x0 + εx1 + . . . ) + 3(1 + ε)(x0 + εx1 + . . . ) + 2(x0 + εx1 + . . . ) = 0, with x(0) = 2(1 + ε), x(0) = −3(1 + 2ε)

Combining the sums with the same power of ε, and recognizing that all orders of ε are linearly independent, sincethe of all powers of ε must be zero, each order of ε sums to zero, ie,

ε0 : x′′0(t) + 3x′0(t) + 2x0(t) = 0, x0(0) = 2, x′0(1) = −3

ε1 : x′′1(t) + 3x′1(t) + 2x1(t) + 3x′0 = 0, x1(0) = 2, x′1(t) = −6

Solving these sequentially (note that since the initial conditions are of mixed order of ε they are separated as well),

x0 = e−2t(et + 1

)x1 = e−2t

(−6t+ et(3t+ 1) + 1

)We find the exact and the two term approximate solution to Equation 40 to be,

xAppx = e−2t(−6tε+ et(3tε+ ε+ 1) + ε+ 1

)Plotting the cases of ε = 0.1, ε = 0.25 and ε = 0.5.


0.2 0.4 0.6 0.8 1.0t

1.0

1.5

2.0

x

Exact

Approximation

Figure 2: ε = 0.1

0.2 0.4 0.6 0.8 1.0t

1.0

1.5

2.0

2.5

x

Exact

Approximation

Figure 3: ε = 0.25

0.2 0.4 0.6 0.8 1.0t

1.0

1.5

2.0

2.5

3.0

x

Exact

Approximation

Figure 4: ε = 0.5



Find the solution of the transcendental equation

sinx = ε cos 2x, (41)

near x = π for small positive ε.If we substitute x = x0 + εx1 + ε2x2 + . . . into Equation 41, we find

sin (x0 + εx1 + ε2x2 + . . . ) = ε cos 2(x0 + εx1 + ε2x2 + . . . ) (42)

Performing a Taylor Series on Equation 42 about ε = 0 and collecting by powers of ε,

ε0 : sin(x0) = 0 (43)

ε1 : cos(x0)x1 − cos(2x0) = 0

ε2 : −1

2sin(x0)x21 + 2 sin(2x0)x1 + cos(x0)x2 = 0

ε3 : −1

6cos(x0)x31 + 2 cos(2x0)x21 − sin(x0)x2x1 + 2 sin(2x0)x2 + cos(x0)x3 = 0

......

Solving Equation 43, we see x0 = π, x1 = −1, x2 = 0, x3 = 116 . Therefore

x = π − ε+11

6ε3 + . . .

problem 1 - nd.edupowers/ame.60611/hwex.13/hw4.sol.pdf · examining this problem, we see that as...

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